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Essential Calculus: Early Transcendentals
Found in: Page 443
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Determine whether the series is convergent or divergent. If its convergent, find its sum.

\(\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{(n - 1)}}}}} \)

The given geometric series is divergent.

See the step by step solution

Step by Step Solution

Rewriting the equation.

\(\begin{aligned}{l}\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {{3^n}{e^{{{(n - 1)}^{ - 1}}}}} \\\end{aligned}\)

\( = \sum\limits_{n = 1}^\infty {{3^n}{e^{(1 - n)}}} \)

Writing in terms of b and q.

\(\sum\limits_{n = 1}^\infty {{3^n}{e^{(1 - n)}}} \) is a geometric series with first term b = \(\frac{3}{e}\)and common ratio q = \(\frac{3}{e}\).

Applying ratio test.

Since \(\left| q \right| = \left| {\frac{3}{e}} \right| > 1\), then by ratio test the series is divergent.

Hence, \(\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{n - 1}}}}} \) is divergent.

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